document.write( "Question 1018556: Three solutions contain a certain acid. The first contains 10% acid, the second 30%, and the third 50%. A chemist wishes to use all three solutions to obtain a 100-liter mixture containing 20% acid. If the chemist wants to use twice as much of the 50% solution as of the 30% solution, how many liters of each solution should be used?
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document.write( " L of the 10% solution
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document.write( " L of the 30% solution
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document.write( " L of the 50% solution
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Algebra.Com's Answer #634634 by Boreal(15235) You can put this solution on YOUR website! x=10%, y=30%, z=50% \n" ); document.write( "x+y+z=100 \n" ); document.write( ".10x+.30y+.50z=20, which is 100*0.20 or the amount of \"pure acid\". \n" ); document.write( "But we know that z=2y \n" ); document.write( "Therefore x+3y=100 \n" ); document.write( "and .10x +.30y+.50(2y)=20 \n" ); document.write( "multiply second by 10 \n" ); document.write( "x+3y+10y=200 \n" ); document.write( "x=100-3y \n" ); document.write( "substitute back \n" ); document.write( "100-3y+3y+10y=200 \n" ); document.write( "10y=100 \n" ); document.write( "y=10 liters \n" ); document.write( "z=20 liters \n" ); document.write( "x=70 liters \n" ); document.write( "10%*70 liters is 7 liters of pure acid \n" ); document.write( "30% of 10 liters is 3 liters \n" ); document.write( "50% of 20 liters is 10 liters \n" ); document.write( "They add to 20 liters \n" ); document.write( "10%:70 liters \n" ); document.write( "30%: 10 liters \n" ); document.write( "50%: 20 liters \n" ); document.write( " |